y(x)由方程y=tan(x+y)确定,求y''

来源：学生作业帮编辑：搜狗做题网作业帮分类：数学作业时间：2024/04/29 07:31:18

y = tan(x + y)
y' = (x+y)'/cos²(x+y) = (1 + y')/cos²(x+y)
y'cos²(x+y) = 1 + y'
[cos²(x+y) - 1]y' = 1
[cos²(x+y) - sin²(x+y) - cos²(x+y)]y' = 1
y' = -1/sin²(x+y)
y'' = -1*(-2)[sin(x+y)]'/sin³(x+y)
= 2cos(x+y)(x+y)'/sin³(x+y)
= 2cos(x+y)(1 + y')/sin³(x+y)
= 2cos(x+y)[1 -1/sin²(x+y)]/sin³(x+y)
= 2cos(x+y)[sin²(x+y) -1]/[sin(x+y)]^5
= 2cos(x+y)[sin²(x+y) - sin²(x+y) - cos²(x+y)]/[sin(x+y)]^5
= -2cos³(x+y)/[sin(x+y)]^5 (^5:5次方)

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